---
title: "Leibnitz theorem: log y = tan⁻¹x"
subject: "Mathematics I"
chapter: 1
key: "semester_1/mathematics_i:leibnitz-log-atan"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/leibnitz-log-atan
site: BCT Engineering
---

# Leibnitz theorem: log y = tan⁻¹x

Mathematics I · Chapter 1: Derivatives and their Applications

- Asked in: 2074 Chaitra, 1+4 marks
- Video: [If y=e^tan^-1(x),prove that  (1+x^2)Yn+2+(2(n+1)x-1)Yn+1 + n(n+1)Yn=0{based on Leibnitz Theorem}](https://www.youtube.com/watch?v=rtDu5QW0geY) — KSB Maths
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/leibnitz-log-atan
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-1

## Question

State Leibnitz theorem. If log y = tan⁻¹x, then show that (1+x²)y_{n+2} + (2nx+2x−1)y_{n+1} + (n²+n)y_n = 0.

## Short answer

Leibnitz theorem: (uv)_n = ∑ C(n,r) u_{n-r} v_r. For log y = tan⁻¹x, (1+x²)y₁ = y. Applying Leibnitz n+1 times gives (1+x²)y_{n+2} + (2nx+2x-1)y_{n+1} + (n²+n)y_n = 0.

## Answer

**Leibnitz's theorem.** If u and v are functions of x, the nth derivative of the product uv is

```text
(uv)ₙ = uₙv + C(n,1)uₙ₋₁v₁ + C(n,2)uₙ₋₂v₂ + ... + u vₙ
```
where C(n,r) = n!/(r!(n−r)!).

#### Proof
log y = tan⁻¹x. Differentiate: y₁/y = 1/(1 + x²), so

**(1 + x²) y₁ = y** …(1)

Differentiate (1) (n+1) times by Leibnitz's theorem, with u = y₁ and v = 1 + x² (v₁ = 2x, v₂ = 2):

```text
(1+x²)yₙ₊₂ + (n+1)(2x)yₙ₊₁ + [(n+1)n/2](2)yₙ = yₙ₊₁
(1+x²)yₙ₊₂ + (2nx + 2x − 1)yₙ₊₁ + (n² + n)yₙ = 0
```
**Proved.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. From log y = tan^-1 x we get:
   A. (1+x^2) y_1 = y
   B. (1-x^2) y_1 = y
   C. (1+x^2) y_1 = 1
   D. y_1 = y
   Answer: A. y_1/y = 1/(1+x^2). (Key point: Base relation)

2. The relation is differentiated:
   A. n+1 times
   B. 2n times
   C. once
   D. n times
   Answer: A. To get y_{n+2} from y_1. (Key point: Differentiations)

3. The (n+1)th derivative of y on the right gives:
   A. y
   B. (n+1) y_n
   C. y_n
   D. y_{n+1}
   Answer: D. d^(n+1)/dx^(n+1) of y is y_{n+1}. (Key point: RHS)

4. Coefficient of y_n is:
   A. n^2 + n
   B. n^2
   C. n^2 - n
   D. 2n
   Answer: A. [(n+1)n/2](2) = n(n+1). (Key point: y_n coefficient)

## Close variants set in other exams

- [Leibnitz theorem: y = e^(x²)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aleibnitz-exp-x-squared) — Mathematics I, chapter 1, asked 2×

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